# Free Spectral Range in wavelength

1. Jan 28, 2012

### Niles

1. The problem statement, all variables and given/known data
Hi

The FSR of a laser with length L is given by Δv = c/2L. If I want to find what Δv is in wavelength, then how do I do this? Obviously I need c=vλ, from which we know that Δv/v = Δλ/λ. But the last equation I can't apply to this case, since it refers to errors around some λ/v.

Best,
Niles.

2. Jan 28, 2012

### SammyS

Staff Emeritus
Errors?

What errors are you referring to?

3. Jan 28, 2012

### Niles

For example if I have a signal at λ with some width Δλ, then I can find the corresponding width Δv in frequency. But you say that it is applicable? Say that FSR = 3GHz, how would I use Δv/v = Δλ/λ to find Δλ?

4. Jan 28, 2012

### SammyS

Staff Emeritus
You would need to know either λ or ν .

5. Jan 28, 2012

### Niles

So you are telling me that the FSR in wavelength-space is not constant? I don't understand how that can be possible. When I say a cavity has a FSR of e.g. 300 MHz, it means the cavity modes are spaced by 300 MHz regardless of the frequency of the modes. So the spacing between mode m and m+1 is 300 MHz.

Now say I want to find the spacing between mode m and m+1 in wavelength. What do I do? Say I know that m is at X GHz and m+1 at Y GHz.

6. Jan 28, 2012

### SammyS

Staff Emeritus
If $\displaystyle\frac{\Delta\nu}{\nu}$ is relatively small (and this implies that m is relatively large.), then $\displaystyle\frac{\Delta\nu}{\nu}\approx\frac{ \Delta\lambda}{\lambda}\,.$ And in this case the wavelengths of successive modes are separated by a nearly constant amount.

Take your example of a cavity with FSR (Free Spectral Range) = 300 MHz:
For m = 1, 2, 3, 4; the frequencies are 300, 600, 900, and 1200 MHz, respectively. I.e. they're multiples of 1, 2, 3, 4 time the fundamental cavity frequency. However, the wavelength of the fundamental is 0.9993 meters ≈ 1 m . The wavelengths for m = 1, 2, 3, 4; are 1 m, 0.5 m, 0.3333 m, 0.250 m . These are nowhere near being equally spaced.

However, consider m = 998, 999, 1000, 1001, 1002; The wavelengths for these modes are 1.0013 mm, 1.0003 mm, 0.9993 mm, 0,9983 mm, and 0.9973 mm. Looking at the difference in wavelength from one mode to the next we find 1.0023×10-3 mm, 1.0003×10-3 mm, 0.9983×10-3 mm, 0.99631×10-3 mm.
The spacing changes by about 0.2% per mode when m is near 1000.

7. Jan 28, 2012

### Niles

Thanks for that, now it is a little more clear. Can I ask how one derives Δv/v = Δλ/λ? As far as I can see from your post, it is only an approximation?

OK, so now I am looking at wavelengths around 1000nm, and I want to find the mode spacing here. The FSR is still 300MHz. So I use Δv/v = Δλ/λ, and we get λ2Δv/c = Δλ. The LHS is

(1000 nm)2 * 300 MHz/c

So this is the FSR in wavelength around 1000nm?

Best,
Niles.